\(\dfrac{1}{2}+\dfrac{1}{4}\\ =\dfrac{2}{4} + \dfrac{1}{4}\\ = \dfrac{3}{4}\)
Same as Mellie's answer.
This could not be the same.
\(\left(\dfrac{a}{b}\right)^{m\cdot n}\)
\(=\left(\dfrac{a^m}{b^m}\right)^n\)
This is correct, this also equals
\(\dfrac{a^{mn}}{b^{mn}}\)
\(\text{This could not be further simplified.}\)
y = exp(x) = ex
Where e = \(1+\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!}......\)
Which is Euler's number
\(\begin{array}{rl}&28-(-43)\\=&28+43\\=&71\end{array}\)
\(\begin{array}{rl}&-28-(-43)\\=&43-28\\=&15\end{array}\)
That is the calculator method.
How to work out 8^(5/3) manually.
\(8^{\frac{5}{3}}=(\sqrt[3]{8})^5=2^5=2\cdot 2 \cdot 2 \cdot 2 \cdot 2 = 4 \cdot 2 \cdot 2 \cdot 2 = 8 \cdot 2 \cdot 2 = 16 \cdot 2 = 32\\ \text{Or if you know what's 2^5, the answer is straight-foward-ly 32.}\)
See your latest post. Melody replied to your post about the derivative one.
Good job Guest #2
\(\dfrac{d \;f(x)}{dx}=1x^{-1}\\ \dfrac{d}{dx} f(x)=\dfrac{1}{x}\\ f(x) = \int \dfrac{dx}{x}\\\quad\;\;\;\;\!=\ln|x|\)
\(\dfrac{d}{dx} 9\ln|x|=\dfrac{9}{x}\)
Melody's way is a good way but I never knew this. Thanks Melody for presenting this method.
